So why exactly do you believe it's false? Anyway, hint: Assume there is an element in A that is not in C... (It might also help to visualize the two products as e.g. rectangles in $\mathbb{R}^2$.)
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anonSep 27 '11 at 10:44

@BleuCheese I've added LaTeXed version of your formula. If you are satisfied with the result, you can edit the post ant leave the LaTeX-ed version there. (Or make further edits, if needed.) For more about writing math at this site see here or here.
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Martin SleziakJun 27 '12 at 14:13

1 Answer
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Well this is true if A and B are not empty sets and false in general.
If $A=\emptyset$, $B=\{1,2\}$, $C=\{1\}$, $D=\{2\}$ then $B\not\subseteq D$ but $A\times B=\emptyset\subseteq C\times D$.

If $A$ and $B$ are not empty the proof is simple enough.

If $A\times B \subseteq C\times D$ then $\forall a\in A, b\in B, (a,b)\in C\times D$ and so $\forall a\in A, a\in C$ and $\forall b\in B, b\in D$ so $A\subseteq C$ and $B\subseteq C$. This is a rather informal sketch but you should figure out where we need the assumption that A and B are not empty sets.